Chapter 16 Additional Aspects of Aqueous Equilibria. The Common Ion Effect and Buffers A buffer is a combination of a weak acid or base with its salt.

Transcription

1 Chapter 16 Additional Aspects of Aqueous Equilibria The Common Ion Effect and Buffers A buffer is a combination of a weak acid or base with its salt. 0.1 M HOAc and 0.1 M NaOAc HOAc H + + OAc Addition of the common ion of a weak acid, in this case its conjugate base causes a shift in the equilibrium. [ OAc ] K a = H+ [ HOAc] OAc [ H + ] = K a HOAc Reaction if add base OH ( aq) + HA( aq ) A ( aq) + H 2 O ( l) Reaction if add acid H + aq ( ) + A aq ( ) HA aq ( ) Chapter 16 Addn. Aspects of Aq. Equil. 1

2 ph of a Buffer [ OAc ] K a = H+ [ HOAc] OAc [ H + ] = K a HOAc log( [ H + ]) = log K a HOAc [ OAc ] log( [ H + ]) = ph log K a = pk a [ ph = pk a log HOAc ] OAc = pk a + log OAc [ HOAc] General form of Henderson-Hasselbalch Equation [ ph = pk a + log base ] [ acid] Example: Calculate the ph of a buffer that is 0.10 M HOAc and 0.20 M NaOAc. Chapter 16 Addn. Aspects of Aq. Equil. 2

3 Maybe easier OAc [ H + ] = K a HOAc ph = log K a [ HOAc] OAc One defining characteristic of a buffer is its ability to resist changes in ph upon the addition of small amounts of either acid or base. When adding acid to buffer H + ( aq) + A ( aq) HA( aq ) mol HA + mol acid added ph = log K a mol A mol acid added When adding base OH ( aq) + HA( aq ) A ( aq) + H 2 O ( l) mol HA - mol base added ph = log K a mol A + mol base added Example: 1. Calculate the ph of a solution, which is 0.25 M in HOAc and 0.25 M in NaOAc. 2. Calculate the resulting ph if 5.0 ml of 0.25 M HCl is added to 100 ml of the above solution. 3. Calculate the resulting ph if 5.0 ml of 0.25 M NaOH is added to 100 ml of the solution in problem Do problem 3 with 50.0 ml of 0.25 M NaOH instead of 5 ml. Chapter 16 Addn. Aspects of Aq. Equil. 3

4 Chapter 16 Addn. Aspects of Aq. Equil. 4

5 Buffer Capacity Is the amount of acid or base that can be added before the buffer loses its ability to resist the change in ph. A more concentrated buffer can react with more added acid or base than a less concentrated one. A buffer acts effectively within a range of ±1 units of pk a. Preparing a buffer solution with a specific ph If the molar concentrations of the acid and conjugate base are about the same, then [ conjugate base] log [ acid] 0 then, ph pk a a. Choose a weak acid whose pk a is close to the desired ph. b. Then substitute the ph and pka values into the Henderson-Hasselbalch equation to obtain the ratio [conjugate base]/[acid]. Calculate the ratio of the molarities of CO 3 2- and HCO 3 - ions required to achieve buffering at ph = the pk a2 = Chapter 16 Addn. Aspects of Aq. Equil. 5

6 Titrations Titration is a procedure for determining the concentration of a solution using another solution of known concentration, called a standard solution. Equivalence point: the number of moles of base (acid) added equals the number of moles of acid (base) At equivalence point, must consider the possibility of hydrolysis. Almost always will be adding a strong acid (base). Strong acid with strong base: At equivalence, the solution is salt of strong acid and strong base, ph = 7 Strong Acid Strong Base Chapter 16 Addn. Aspects of Aq. Equil. 6

7 Weak acid with strong base Upon addition of base the solution can be treated like an acidic buffer. At equivalence, the solution is salt of weak acid and strong base: basic: ph > 7 Strong acid with weak base: ph < 7 Upon addition of base the solution can be treated like an basic buffer. Solution is salt of strong acid and weak base: acidic. Chapter 16 Addn. Aspects of Aq. Equil. 7

8 Acid Base Indicators Indicators usually have distinctly different colors in their nonionized and ionized forms. These two forms are related to the ph of the solution. The end point of a titration occurs when the indicator changes color. HIn(aq) H + (aq) + In - (aq) [ HIn [ In 10 color of acid (HIn) predominates ] HIn In 0.1 color of base (In ) predominates Chapter 16 Addn. Aspects of Aq. Equil. 8

9 Precipitation Reactions, Solubility Products, Ksp Soluble Compounds Almost all salts of Na +, K +, and NH 4 + Exceptions All salts of Cl -, Br -, and I - Halides of Ag +, Hg 2 2+, and Pb 2+ Compounds containing F- Fluorides of Mg 2+, Ca 2+, Sr 2+, Ba 2+, Pb 2+ Salts of nitrate, NO 3 - chlorate, ClO 3 - perchlorate, ClO 4 - acetate, CH 3 CO 2 - Salts of sulfate, SO 4 2- Sulfates of Sr 2+, Ba 2+, Pb 2+ Insoluble Compounds 2- All salts of carbonates, CO 3 3- phosphates, PO 4 2- oxalate, C 2 O 4 2- chromate, CrO 4 sulfide, S 2- Most metal hydroxides and oxides Exceptions Salts of NH 4 + and the alkali metal cations Chapter 16 Addn. Aspects of Aq. Equil. 9

10 The Solubility Product Constant, Ksp Ag + (aq) + Br (aq) AgBr (s) AgBr (s) Ag + (aq) + Br (aq) K sp = [ Ag + (aq)][ Br (aq)] = at 25 C In the case above [ Ag + (aq)] = Br (aq) = = M Generic A x B y (s) xa y+ (aq) + yb x- (aq) K sp = [ A y + ] x [ B x ] y Example: Pb 3 (PO 4 ) 2 Chapter 16 Addn. Aspects of Aq. Equil. 10

11 Precipitation Reactions insoluble product precipitate Net Ionic Reactions only involves ions that form precipitate spectator ions don't precipitate; not in net ionic reactions Estimating Salt Solubility From K sp 1. Write the K sp expression. 2. Let x = molar solubility of the salt 3. Use stoichiometry of reaction to express the concentration of each species in terms of x. 4. Substitute these concentrations into the equilibrium expression and solve for x Compound K sp expression Cation Anion Relationship between K sp and s AgCl [Ag + ][Cl - ] s s K sp =s 2 BaSO 4 [Ba 2+ ][ SO 4 2 ] s s K sp =s 2 Ag 2 CO 3 [Ag + ] 2 [CO 3 2- ] 2s s K sp =4s 3 PbF 2 [Pb 2+ ][F - ] 2 s 2s K sp =4s 3 Al(OH) 3 [Al 3+ ][OH - ] 3 s 3s K sp =27s 4 Ca 3 (PO 4 ) 2 [Ca 2+ ] 3 [ PO 4 3- ] 2 3s 2s K sp =108s 5 Chapter 16 Addn. Aspects of Aq. Equil. 11

12 Determining K sp from solubility 1. Convert solubility to molar solubility 2. Convert molar solubility to concentration of ions at equilibrium. 3. Use equilibrium concentrations of ions in K sp expression. The K sp for Ba(IO 3 ) 2 is 6.0x10 10 at 25 C. Calculate the molar solubility of Ba(IO 3 ) 2 and the solubility in grams per liter. The molar solubility of PbBr 2 at 25 C is 1.0x10-2 mol/l. Calculate K sp. Chapter 16 Addn. Aspects of Aq. Equil. 12

13 Factors that Affect Solubility Solubility and the Common Ion Effect Solubility and ph Complex Ion Formation and Simultaneous Equilibria Common Ion Solubility is decreased when a common ion is added. CaF 2 ( s) = Ca 2+ ( aq)+ 2F ( aq) Adding F will cause equilibrium to shift left. CaF 2 (s) will form and precipitation occurs. As NaF is added to the system, the solubility of CaF 2 decreases. Calculate the solubility of CaF 2 in (a) pure water; (b) 0.15 M KF solution; (c) M Ca(NO 3 ) 2 solution. K sp = 3.9x Chapter 16 Addn. Aspects of Aq. Equil. 13

14 Solubility and ph Mg( OH) 2 ( s) = Mg 2+ ( aq) + 2OH ( aq) If OH is removed from solution, equilibrium shifts right and the solubility increases. OH can be removed from solution by adding strong acid. OH aq ( ) + H + ( aq) = H 2 O l ( ) As ph decreases, [H + ] increases, solubility increases CaF 2 ( s) = Ca 2+ ( aq)+ 2F aq ( ) Removing F (aq) causes equilibrium to shift right and increase solubility. 1. Add strong acid ( ) + F ( aq) = HF( aq) H + aq As ph decreases, [H + ] increases, solubility of CaF 2 increases. Effect is most significant if one or both ions involved are somewhat acidic or basic. (OH, F, OAc, CO 3 2-, S 2- ) Complex Ions AgCl, K sp = Solubility increases if NH 3 is added. Ag + + ( aq)+ 2 NH 3 ( aq) = Ag( NH 3 ) 2 ( aq) Chapter 16 Addn. Aspects of Aq. Equil. 14

15 Constant for reaction is called formation constant, K f + [ ( ) 2 ] K f = Ag NH 3 Ag + NH 3 2 = Addition of NH 3 to AgCl (white precipitate) Net result is that Ag + is removed from solution. K overall = K sp x K f Calculate the concentration of Ag + present when concentrated ammonia is added to M solution of AgNO 3 to give an equilibrium concentration of [NH 3 ] = 0.20 M. Neglect the small volume change that occurs on addition of NH 3. Chapter 16 Addn. Aspects of Aq. Equil. 15

16 Solubility, Ion Separations, and Qualitative Analysis BaSO4( s) = Ba 2+ 2 ( aq) + SO 4 ( aq) Q = [ Ba 2+ 2 ][ SO 4 ] If: Q > K sp ppt forms until Q = K sp Q = K sp Q < K sp equilibrium exists solid dissolves until Q = K sp Selective Precipitation If add HCl to a solution of Ag + and Cu 2+, silver ppt (K sp AgCl = 1.8 x ) while Cu 2+ remains in solution. Removal of one metal from solution: selective precipitation. Ex: Zn 2+ and Cu 2+ CuS (K sp = 6 x ) is less soluble than ZnS (K sp = 2 x ) Cu 2+ will be removed first As add H 2 S to green solution, black CuS ppt forms Zn 2+ (colorless solution) remains As add more H 2 S, second ppt of white ZnS forms. Chapter 16 Addn. Aspects of Aq. Equil. 16